## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 11 - Section 11.2 - Arithmetic and Geometric Sequences - Exercise Set: 8

#### Answer

$1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}$

#### Work Step by Step

The general term $a_n$ of a geometric sequence is given by $a_n = a_1r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio. We plug in $n = 1, 2, 3, 4, 5$ to find the first five terms of the geometric sequence $a_1 = 1\cdot\frac{1}{3}^{1-1} = 1\cdot\frac{1}{3}^0=1\cdot1=1$ $a_2 = 1\cdot\frac{1}{3}^{2-1} = 1\cdot\frac{1}{3}^1=1\cdot\frac{1}{3}=\frac{1}{3}$ $a_3 = 1\cdot\frac{1}{3}^{3-1} = 1\cdot\frac{1}{3}^2=1\cdot\frac{1}{9}=\frac{1}{9}$ $a_4 = 1\cdot\frac{1}{3}^{4-1} = 1\cdot\frac{1}{3}^3=1\cdot\frac{1}{27}=\frac{1}{27}$ $a_5 = 1\cdot\frac{1}{3}^{5-1} = 1\cdot\frac{1}{3}^4=1\cdot\frac{1}{81}=\frac{1}{81}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.